sequences and series - Show $lim_{Nto infty}sum_{k=1}^{N}frac{1}{k+N}=ln(2)$

I have some difficulty to prove the following limit:
$$\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$$
Can someone help me? Thanks.

Answer

A common estimate for the Harmonic Numbers is
$$
\sum_{k=1}^n\frac1k=\log(n)+\gamma+O\left(\frac1n\right)\tag{1}

$$
where $\gamma$ is the Euler-Mascheroni constant.

Applying $(1)$, we get that
$$
\begin{align}
\sum_{k=1}^{N}\frac{1}{k+N}
&=\sum_{k=1}^{2N}\frac1k-\sum_{k=1}^N\frac1k\\
&=\left(\log(2N)+\gamma+O\left(\frac{1}{2N}\right)\right)-\left(\log(N)+\gamma+O\left(\frac1N\right)\right)\\
&=\log(2)+O\left(\frac1N\right)\tag{2}

\end{align}
$$
Taking the limit of $(2)$ as $N\to\infty$ yields
$$
\lim_{N\to\infty}\sum_{k=1}^{N}\frac{1}{k+N}=\log(2)\tag{3}
$$